How Fast Will the Skier Be After Crossing the Snow Patch?

[Fera09]

Homework Statement

Part A. A 64.0-kg skier starts from rest at the top of a ski slope of height 70.0 m.
If frictional forces do −1.00×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
I already got the right answer for this one.. I have problems in this second part.

Part B. Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.19. If the patch is of width 69.0 m and the average force of air resistance on the skier is 140 N, how fast is she going after crossing the patch?

Homework Equations

V_initial= 32.5 (got it from part A)
V_final= ?
Distance = 69.0 m
F_normal= mg = 627.84N
coefficient of friction = 0.19
m= 64.0 kg

F_friction= F_normal * coefficient of friction

W_friction= -F_friction*Distance
W_{airresistance}= -F_{airresistance}*Distance (is this right?)
Change in kinetic energy= (1/2)mv_final²-(1/2)mv_initial²

The Attempt at a Solution

So I got the equation..
W_friction+W_{airresistance}=(1/2)mv_final²-(1/2)mvi_initial²

and i plugged in..

-8222.59 + -9660 = (1/2)(64)v_final²-(1/2)(64)(32.5)


I know I'm doing something wrong somewhere because I can't take the square root of a negative number, but i don't know where.

 

[Fera09]

just noticed i put this in the wrong section, sorry!

 

[berkeman]

just noticed i put this in the wrong section, sorry!

No, I moved it here from Advanced Physics. This belongs here in Intro Physics. Check your PMs.

 

[alphysicist]

Hi Fera09,

Homework Statement

Part A. A 64.0-kg skier starts from rest at the top of a ski slope of height 70.0 m.
If frictional forces do −1.00×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
I already got the right answer for this one.. I have problems in this second part.

Part B. Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.19. If the patch is of width 69.0 m and the average force of air resistance on the skier is 140 N, how fast is she going after crossing the patch?

Homework Equations

V_initial= 32.5 (got it from part A)
V_final= ?
Distance = 69.0 m
F_normal= mg = 627.84N
coefficient of friction = 0.19
m= 64.0 kg

F_friction= F_normal * coefficient of friction

W_friction= -F_friction*Distance
W_{airresistance}= -F_{airresistance}*Distance (is this right?)
Change in kinetic energy= (1/2)mv_final²-(1/2)mv_initial²

The Attempt at a Solution

So I got the equation..
W_friction+W_{airresistance}=(1/2)mv_final²-(1/2)mvi_initial²

and i plugged in..

-8222.59 + -9660 = (1/2)(64)v_final²-(1/2)(64)(32.5)


I know I'm doing something wrong somewhere because I can't take the square root of a negative number, but i don't know where.

In your last line, did you forget to square the 32.5 (or is that just a typing error)?